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Calculating the Optical Rotation Value

by Interception
Tags: optical, rotation
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Interception
#1
Nov29-12, 06:03 PM
P: 15
1. The problem statement, all variables and given/known data So the problem states that we have a (+) enantiomer of a coumpund with an optical rotation of 50*. If a pure sample containes 42% of the (+) enantiomer and 58% of the (-) enantioimer, what is the optical rotation value. (By * i mean degrees)



2. Relevant equations I'm not sure which equation would apply to this situation. It's a small school, and our proffesor hardly went over this subject. We'd done a few problems, but none where like this.



3. The attempt at a solution I know that it will be a negative value since the the (-) enantiomer outnumbers the (+). I didn't know how to approach the problem, so I took 58%-the 42% to see how much of the (-) will contribute to the rotations since the 42% of the (+) enantiomer will counteract 42% of the (-) enantiomer. I got -16. I took this mean that 16% of the (-) enantiomer would be optically active, and 16% of 50 is 8, or in this case -8* of optical rotation. The negative value makes sense, and the fact that it's not a large number. Does this look right?
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Borek
#2
Nov30-12, 01:53 AM
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Borek's Avatar
P: 23,363
Looks OK to me.


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